Answer is:
The sun's energy is transferred through the vacuum of space to Earth
<h3><u>Answer;</u></h3>
Empirical formula = C₂H₃O
Molecular formula = C₁₄H₂₁O₇
<h3><u>Explanation</u>;</h3>
Empirical formula
Moles of;
Carbon = 55.8 /12 = 4.65 moles
Hydrogen = 7.04/ 1 = 7.04 moles
Oxygen = 37.16/ 16 = 2.3225 moles
We then get the mole ratio;
4.65/2.3225 = 2.0
7.04/2.3225 = 3.0
2.3225/2.3225 = 1.0
Therefore;
The empirical formula = <u>C₂H₃O</u>
Molecular formula;
(C2H3O)n = 301.35 g
(12 ×2 + 3× 1 + 16×1)n = 301.35
43n = 301.35
n = 7
Therefore;
Molecular formula = (C2H3O)7
<u> = C₁₄H₂₁O₇</u>
<span>during photosynthesis the co2 is converted into sugar
</span><span> 6 carbon compound that immediately splits into 2 molecules of 3-phosphoglycerate.
</span><span>3 CO2 + 9 ATP + 6 NADPH + 6 H+ → C3H6O3-phosphate + 9 ADP + 8Pi + 6 NADP+ + 3 H2O
</span>hope it helps
Answer:
4KO₂ + 2CO₂ -> 2K₂CO₃ + 3O₂
<u> Step 1: Find the moles of O₂.</u>
n(O₂) = mass/ Mr.
n(O₂) = 100 / 32 = 3.125 mol
<u>Step 2: Find the ratio between KO₂ and O₂.</u>
<u>KO₂ </u> : <u> O₂</u>
4 : 3
4/3 : 1
(4*3125)/3 : 3.125
=4.167 mol of KO₂
Thus now we know, to produce 100 g of O₂, we need 4.167mol of KO₂
<u>Step 3: Find the mass of KO₂:</u>
<u />
mass = mol * Mr. (KO₂)
Mass = 4.167* 71.1
Mass = 296.25 g
Answer : The mass of ethylene glycol is, 437.34 grams
Solution : Given,
Volume of ethylene glycol = 394 ml
Density of ethylene glycol = 1.11 g/ml (Standard value)
Formula used :
Now put all the given values in this formula, we get the mass of ethylene glycol.
Therefore, the mass of ethylene glycol is, 437.34 grams